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3 years ago

Alice tried to solve the equation. What was her mistake?

2 answers:

8 0

She didn’t multiply -1/4 by 12 in order to get -1/4x + 3 = -1

which would change her answer, because then you have to subtract 3 from -1 and get -4. Then divide -1/4 & get 16. It changed to positive because of the division & multiplication rule with two negative numbers.

Hope this helps.

Serjik [45]3 years ago

7 0

She forgot to distribute the -1/4 to the -12

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Raina must choose a number between 55 and 101 that is a multiple of 2, 8 and 10. Write all the numbers that she could choose. If

Aleonysh [2.5K]

Answer:

The only possible number is $80$ .

Step-by-step explanation:

The number in question needs to be a multiple of all three of $2$ , $8$ , and $10$ . As a result, it must also be a multiple of the least common multiplier (lcm) of the three number.

Start by finding the least common multiplier of the three numbers.

Factor each number into its prime components:

$2$ is a prime number itself.
$8 = 2^3$ .
$10 = 2 \times 5$ .

The only prime factors are $2$ and $5$ .

The greatest power of $2$ among the three numbers is $3$ .
The greatest power of $5$ among the three numbers is $1$ .

Therefore, the least common multiplier of the three number should be the product of $2^3$ and $5$ . That's equal to $2^3 \times 5 = 8 \times 5 = 40$ .

In other words, the number (or numbers) in question could be written in the form $40\, k$ , where $k$ is an integer.

The question requires that this number be between $55$ and $101$ . In other words,

$55 \le 40\, k \le 101$ .

The goal is to find the possible values of $k$ . Note that from integer division by $40$ ,

$55 = 1 \times 40 + 15$ , and
$101 = 2 \times 40 + 21$ .

The inequality becomes:

$1 \times 40 + 15 \le 40\, k \le 2 \times 40 + 21$ .

However,

$1 \times 40 < 55 = 1 \times 40 + 15$ , and
$2 \times 40 + 21 = 101 < 3 \times 40$ .

Hence,

$1 \times 40 < 1\times 40 + 15 \le 40\, k \le 2 \times 40 + 21 < 3 \times 40$ .

$1 \times 40 < 40\, k < 3 \times 40$ .

Divide by the positive number $40$ to obtain:

$1 < k < 3$ .

Since $k$ is an integers, $k = 2$ .

Indeed, $40 \, k = 80$ is between $55$ and $101$ .

Therefore, $80$ is the number in question.

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4 years ago

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3 years ago

Can one line be shorter than the other and still be parallel

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4 years ago

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under

sweet [91]

Answer: Work done by the particle is 78 N.

Step-by-step explanation:

If C is the path the particle follows, then work done is $\int_{C}^{}F.dx$ .

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therefore, if S is the at surface with boundary C, so that S is the portion of the plane $z=\frac{1}{4}y$ over the rectangle D=[0,2]\times [0,4].

Now,

$curlF=\begin{vmatrix}i &j &k \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ z^{2}&5xy &4y^{2} \end{vmatrix}=8yi+2zj+5yk$

By the Stoke's theorem:

$\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA$

$=\iint_{D}^{}\left ( \frac{39}{8}y \right )dA\, \, \, \, \, \, \, \, \, Since\, \, z=\frac{1}{4}y\\\\=\int_{0}^{2}\int_{0}^{4}\frac{39}{8}y\, \, \, dy\, dx\\\\$